# Why do we use the function derivatives in the Wronskian?

I'm looking for more information on the matrix that the Wronskian acts on to provide information about a set of solutions to an ODE.

So far I understand that:

• The Wronskian comes from Cramer's rule (at least that is the explanation I see most of the time).
• If the Wronskian is not equal to zero, then the set of solutions is linearly independent.
• If the Wronskian is equal to zero, that the set of solutions could still be linearly independent, but likely isn't.
• If the determinant of a matrix is equal to nonzero, then the matrix is linearly independent, it can be inverted and all that jazz.

What I'm confused on:

• Why did we choose the derivatives of the function in the matrix inside of the Wronskian?
• I've seen some... lame proofs that the derivatives of differentiable functions should be "linearly independent" to each other. I can easily think of functions that are differentiable that don't follow that rule (like $$e^x$$).
• (This might be a stupid question): If the columns of the matrix are not linearly independent then isn't the entire matrix linearly dependent? Is this where our problems of the Wronskian doesn't guarantee linear dependence arise from?
• It's not correct terminology to say that a matrix is linearly independent (if what you actually mean is that its columns are). Sep 11 at 7:51
• And what do you mean by “the matrix that the Wronskian acts on”? Sep 11 at 7:52
• You make a fair comment. The Wronskian is defined as the determinant of a matrix where each column is a function and its derivatives. What I guess what I'm asking about is the matrix that is used to calculate the Wronskian. Why do we use a matrix where the columns are the solutions, and their derivatives in the Wronskian? Sep 13 at 17:23
• OK, then it's clear what you mean. Sep 13 at 17:27

$$y(x)=C_1u(x) + C_2v(x)$$ given the initial conditions: $$y(t_0)=y_0$$ and $$y'(t_0)=y'_0$$ means solving for the solution to the matrix below $$\begin{bmatrix} u(t_0)&v(t_0) \\u'(t_0)&v'(t_0) \end{bmatrix} \begin{bmatrix} C_1\\C_2 \end{bmatrix} = \begin{bmatrix} y_0\\y'_0 \end{bmatrix}$$
In the interval $$(a,b)$$, the complete solution to the higher order linear differential equation must have $$u,v$$ to be linearly independent on $$(a,b)$$, the Wronskian not to be zero at some point $$t_0$$ in $$(a,b)$$ or never zero at all. If one of these conditions are met, the solution is called a fundamental solution set on the ODE at $$(a,b)$$.
• I realized that it is important to note that we've assumed $u$ and $v$ are solutions of the ODE. Sometimes textbooks will say you can use the Wronskian on any set of functions, but the Wronskian is defined for a set of solutions. Thus if the Wronskian equals zero, then the set of solutions is linearly dependent. Sep 13 at 23:10